We denote by C a positive absolute constant, which may change from line
to line. We also denote by A1, A2,... positive constants that depend only on the
data. Here we need to use a bootstrap argument to get the decay estimate of \(\theta ^n\) .
Step 1. An auxiliary estimate
We write the temperature equation of (REF ) as the following integral equation
\(\theta ^n(t)=e^{t\Delta }\theta _0+\int _0^te^{(t-s)\Delta }\Psi _\delta (u^{n-1})\nabla \theta ^n ds+\int _0^te^{(t-s)\Delta }\nabla (a^n\nabla \theta ^n)ds, \) 

where
\(e^{t\Delta }f(x)=\int _{R^3}E_t(x-y)f(y)dy,\ E_t(x)=\frac{1}{(4\pi t)^{\frac{3}{2}}}e^{-\frac{|x|^2}{4t}}.\) 
We follow the Fourier splitting method introduced in [1]}, [2]}. Using the Plancherel theorem in the energy inequality for \(\theta ^n\) , we get
\(\frac{1}{2}\frac{d}{dt}\int |\widehat{\theta ^n(\xi ,t)}|^2d\xi \le -c_0\int |\xi |^2|\widehat{\theta ^n(\xi ,t)}|^2d\xi ,\) 
then split the integral on the right-hand side into \(S\cup S^c\) , where
\(S=\lbrace \xi :|\xi |\le (\frac{k}{2c_0(t+1)})^{\frac{1}{2}} \rbrace \) 
it follows that
\(\frac{1}{2}\frac{d}{dt}\int |\widehat{\theta ^n(\xi ,t)}|^2d\xi +\frac{k}{1+t}\int |\widehat{\theta ^n(\xi ,t)}|^2d\xi \le \frac{k}{1+t}\int _S|\widehat{\theta ^n(\xi ,t)}|^2d\xi ,\) 
multiplying by \((1+t)^k\)  we obtain
\(\frac{d}{dt}[\int (1+t)^k|\widehat{\theta ^n(\xi ,t)}|^2d\xi ]\le k(1+t)^{k-1}\int _S|\widehat{\theta ^n(\xi ,t)}|^2d\xi .\) 
Because
\(|\widehat{\theta ^n(\xi ,t)}|^2&\le &2|e^{t\Delta }\theta _0|^2+2|\xi |^2(\int _0^t\Vert u^{n-1}\Vert _2\Vert \theta ^n\Vert _2ds)^2+2|\xi |^2(\int _0^t\Vert a^n\Vert _2\Vert \nabla \theta ^n\Vert _2ds)^2.\) 
Replacing this in (REF ) and applying the Plancherel theorem, we can get
\(&&\frac{d}{dt}\left[\int (1+t)^k\Vert \theta ^n(\xi ,t)\Vert _2^2d\xi \right]\\&&\le C\left[\Vert e^{t\Delta }\theta _0\Vert _2^2(1+t)^{k-1}+(1+t)^{k-\frac{7}{2}}(\int _0^t\Vert u^{n-1}\Vert _2\Vert \theta ^n\Vert _2ds)^2+(1+t)^{k-\frac{7}{2}}(\int _0^t\Vert a^n\Vert _2\Vert \nabla \theta ^n\Vert _2ds)^2\right].\) 
As \(\int \theta _0dx=0\) , \(\theta _0\in L_1^1\) , we easily get
\( \Vert e^{t\Delta }\theta _0\Vert ^2\le A_1(1+t)^{-5/2}.\) 
Let \(k = 7/2\) . For all \(n\in N\) , it follows that,
\(\Vert \theta ^n(t)\Vert _2&\le & C\Big [A_1(1+t)^{-\frac{5}{2}}+(1+t)^{-\frac{7}{2}}\int _0^t(\int _0^s\Vert u^{n-1}(r)\Vert _2\Vert \theta ^n(r)\Vert _2dr)^2ds\nonumber \\&&+(1+t)^{-\frac{7}{2}}\int _0^t(\int _0^t\Vert a^n\Vert _2\Vert \nabla \theta ^n\Vert _2ds)^2\Big ].\) 
Step 2. The inductive argument.
Consider the following induction hypothesis:
\(\Vert u^{n-1}\Vert _2\le C,\quad \int _0^t\Vert \nabla u^{n-1}\Vert _2^2\le C.\) 
For \(n=1\)  the inductive condition (REF ) is immediate since \(u^0= 0\) . Let us now prove that \(\Vert u^{n}\Vert _2\le C\)  and \(\int _0^t\Vert \nabla u^{n}\Vert _2^2\le C\) , assuming that (REF ) holds true. From (REF ), (REF ) and Proposition REF , we have the following estimates,
\(\Vert \theta ^n\Vert _2\le C\Vert \theta _0\Vert _1t^{-\frac{3}{4}},\quad \Vert a^n\Vert _2\le C\Vert a_0\Vert _2,\quad \Vert \nabla \theta ^n\Vert _2\le Ct^{-\frac{5}{4}}.\) 
Putting these inside (REF ), we can obtain
\(\Vert \theta ^n(t)\Vert _2&\le & C\Big [A_1(1+t)^{-\frac{5}{2}}+(1+t)^{-\frac{7}{2}}\Vert \theta _0\Vert _2\int _0^t(\int _0^s\Vert u^{n-1}(r)\Vert _2r^{-\frac{3}{4}}dr)^2ds\nonumber \\&&+(1+t)^{-\frac{7}{2}}\Vert a_0\Vert _2\int _0^t(\int _0^sr^{-\frac{5}{4}}dr)^2ds\Big ].\) 
Then we can get
\(\Vert \theta ^n(t)\Vert _2^2\le (1+t)^{-2}.\) 
Combining the above inequality and following the method of Lemma REF , we have
\(\Vert \nabla \theta ^n\Vert _2\le C(1+t)^{-\frac{3}{2}}.\) 
Inserting the above two inequalities inside (REF ) again, we can obtain
\(\Vert \theta ^n\Vert _2\le C(1+t)^{-\frac{5}{4}}.\) 
Applying the method of Lemma REF  again, we have
\(\Vert \nabla \theta ^n\Vert _2\le C(1+t)^{-\frac{7}{4}}.\) 
Step 3. Uniform bound for the \(L^2\) -norm of the velocities \(u^n\)  and \(\int _0^t\Vert \nabla u^{n}\Vert _2^2\le C\) .
Back to the differential inequality (REF ), we can get
\(\frac{d}{dt}\left\Vert \frac{u^n}{\sqrt{1+a^n}}\right\Vert _2\le \frac{1}{\sqrt{1+m}}\Vert \theta ^n\Vert _2, \) 
As
\(\Vert \theta ^n\Vert _2\le C(1+t)^{-\frac{5}{4}},\) 
we finally get
\(\left\Vert u^n\right\Vert _2\le \left\Vert u_0\right\Vert _2+C\int _0^t(1+s)^{-\frac{5}{4}}ds\le C\) 
Employing the differential inequality (REF ) again, we can get
\(\frac{1}{2(1+M)}\Vert u^n\Vert _2^2 + \int _0^t \Vert \nabla u^n(s)\Vert _2^2 ds&\le &\frac{1}{2}\Vert \frac{u^n}{\sqrt{1+a^n}}\Vert _2^2 + \int _0^t \Vert \nabla u^n(s)\Vert _2^2 ds\\ &\le &\frac{1}{2}\Vert \frac{u_0}{\sqrt{1+a_0}}\Vert _2^2+\int _0^t\int \frac{\theta ^n u^ne_3 }{\sqrt{1+a_0}} dxds\\\\&\le &\frac{1}{2}\Vert \frac{u_0}{\sqrt{1+a_0}}\Vert _2^2+C\int _0^t\Vert \theta ^n\Vert _{2}\Vert u^n\Vert _{2}ds \\&\le &\frac{1}{2(1+m)}\Vert u_0\Vert _2^2+\int _0^t C(1+s)^{-\frac{5}{4}}ds\le C, \ \ for \ \ t>0.\) 
This completes the proof of Lemma REF .
